| 1 | \documentclass{article} |
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| 2 | \usepackage{fullpage} |
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| 3 | \begin{document} |
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| 4 | \begin{enumerate} |
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| 5 | \item Non-Newtonian polymer flow in a channel |
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| 6 | |
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| 7 | A polymer melt is forced through a flat channel of thickness $\delta$ between |
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| 8 | two fixed horizontal plates by a pressure difference $\Delta P$ (inlet |
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| 9 | pressure minus outlet pressure). This channel length $L$ is much longer than |
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| 10 | its width $W$, and it is a lot longer and wider than it is thick ($\delta$), |
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| 11 | so you may assume fully-developed flow and neglect edge effects. Take the |
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| 12 | $x$ direction to be the direction of flow, parallel to the plates, and the |
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| 13 | $y$ direction to be straight up, perpendicular to the plates. |
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| 14 | |
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| 15 | This polymer is a pseudoplastic non-Newtonian fluid, whose behavior can be |
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| 16 | modeled using a ``power law'': |
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| 17 | $$\tau_{yx} = -\mu_0 \left|\frac{\partial u_x}{\partial y}\right|^{n-1} |
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| 18 | \frac{\partial u_x}{\partial y},$$ |
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| 19 | which is to say, shear stress is proportional to shear strain rate to the $n$ |
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| 20 | power, with the extra $\partial u_x/\partial y$ there to get the sign right. |
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| 21 | The differential equation describing $x$-momentum equation for this laminar |
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| 22 | 1-D flow (where $u_y=u_z=0$ and $d\vec{u}/dx=d\vec{u}/dz=0)$ can be written |
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| 23 | with the above power law substituted for $\tau_{yx}$: |
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| 24 | $$\rho\frac{\partial u_x}{\partial t} = -\frac{\partial P}{\partial x} + |
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| 25 | \mu_0\frac{\partial}{\partial y}\left( |
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| 26 | \left|\frac{\partial u_x}{\partial y}\right|^{n-1} |
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| 27 | \frac{\partial u_x}{\partial y}\right) + F_x.$$ |
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| 28 | The general solution to this equation for steady-state flow driven only by |
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| 29 | pressure gradient is: |
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| 30 | $$u_x = -\frac{\mu_0L}{\Delta P}\frac{n}{n+1} |
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| 31 | \left(-\frac{\Delta P}{\mu_0L}y + C_1\right)^{\frac{n+1}{n}} + C_2.$$ |
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| 32 | Note that this is only valid for negative $y$; the positive $y$ solution is |
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| 33 | symmetric. |
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| 34 | |
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| 35 | \begin{enumerate} |
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| 36 | \item If the polymer is pseudoplastic, is $n$ greater or less than one? |
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| 37 | |
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| 38 | \item Determine the specific solution which fits the above general solution |
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| 39 | to the no-slip boundary conditions at the two stationary plates, and/or the |
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| 40 | symmetry plane boundary condition halfway between them (you need just two |
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| 41 | of these three conditions). Note: it might help to set $y=0$ halfway |
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| 42 | between the plates. |
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| 43 | |
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| 44 | \item Sketch this velocity profile for an appropriate pseudoplastic $n$ value |
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| 45 | of your choosing. |
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| 46 | |
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| 47 | \item For channel flow of this kind, a Newtonian fluid's average velocity |
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| 48 | will be 2/3 of its maximum velocity. For this pseudoplastic fluid, will |
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| 49 | the average velocity be more or less than 2/3 of its maximum? |
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| 50 | \end{enumerate} |
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| 51 | \end{enumerate} |
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| 52 | \end{document} |
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